Having trouble coming up with a proof for a problem related to SLLN using MGFS. I found Strong Law of Numbers for $S_{n}$ Bounded Casella Berger 5.38 but the answer doesn't seem to hint at how we could establish some lower bound. I've put my effort below, but it feels cheap, are there other methods in which I can approach this problem?
Problem statement: Let $X_1,...,X_n$ be iid random variables with mgf $M_X(t),-h<t<h$, let $S_n=\sum_{i=1}^nX_i$ and $\bar{X}=\frac{1}{n}S_n$
I've already via Markov inequality shown that $$P(S_n>a)\le e^{-at}[M_X(t)]^n$$ and $$P(S_n \le a) \le e^{-at}[M_X(t)]^n$$ Now we must prove that for $E[X]<0$ and $M_X(0)=1$ there exists $0<c<1$ s.t $P(S_n>a) \le c^n$.
Now my first instinct is to use a Taylor expansion:
$$M_X(t) \approx M_X(0) + M_X'(0)t =1 + tE[X]$$
However I can only see this bringing us to the upperbound in the link mentioned earlier: $$M_X(t) \le 1 + t\frac{E[X]}{2} \le 1$$ Since $E[X]$ is negative, or the second derivative of the mgf is negative near 0. Taking t=0 gives us the 1 as the final upperbound. The nth power of the mgf retains the property of 1 as an upperbound. Since we are confident of the upper-bound on $M_X(t)$ in the neighbourhood of 0 couldn't we just use:
$$P(S_n>a) \le c^n = \left[ e^{-at/n}*M_X(t) \right]^n?$$
We've shown that from $E[X]<0$, $M_X(t)$ is bounded above by 1 in the neighbourhood of 0. In addition as n grows, since $0<c<1$, then $0<c^n<1$. Am I missing something here? I should note that nothing about the nature of $a$ is given, I assumed it was positive above.
Hints would be appreciated, my analysis skill-set is limited and something I'm working on!